A boy on a 2.2 kg skateboard initially at rest
tosses a(n) 7.7 kg jug of water in the forward
direction.
If the jug has a speed of 2.9 m/s relative to
the ground and the boy and skateboard move
in the opposite direction at 0.57 m/s, find the
boy’s mass.
Answer in units of kg
(I just want the formula) due in about 2 hours
Not sure, but I believe it is (M+M)V=MV. Solve for your unknown.
(M+2.2Kg)(0.57m/s)=(7.7Kg)(2.9m/s)
To solve this problem, we can use the principle of conservation of momentum. According to this principle, the total momentum before the jug is thrown should be equal to the total momentum after the jug is thrown.
The formula to calculate momentum is:
Momentum = mass × velocity
Given data:
Mass of the skateboard = 2.2 kg
Mass of the jug = 7.7 kg
Velocity of the jug relative to the ground = 2.9 m/s
Velocity of the skateboard (with the boy) = -0.57 m/s (opposite direction)
Let's assume the mass of the boy on the skateboard as m (in kg).
The total momentum before the jug is thrown is the sum of the momentum of the skateboard (with the boy) and the jug:
Momentum_before = (mass of skateboard + mass of boy) × velocity of skateboard
The total momentum after the jug is thrown is the sum of the momentum of the skateboard (with the boy) and the jug:
Momentum_after = mass of skateboard × velocity of skateboard (after the jug is thrown) + mass of jug × velocity of jug
Since the total momentum before and after the jug is thrown should be equal, we can equate these two expressions:
Momentum_before = Momentum_after
((mass of skateboard + mass of boy) × velocity of skateboard) = (mass of skateboard × velocity of skateboard (after the jug is thrown) + mass of jug × velocity of jug)
Now, we can plug in the given values and solve for the mass of the boy (m):
(2.2 kg + m) × (-0.57 m/s) = (2.2 kg × -0.57 m/s + 7.7 kg × 2.9 m/s)
Solving this equation will give us the mass of the boy.